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Does a passenger jet detect excessive weight in the tail after it's just landed?
My wife flew in on a BA jet yesterday and immediately after landing the pilot told the passengers that the tail was very heavy and that they needed to remove stuff before he'd taxi very far.  I've never come across this before.
Any explanations, please?

and then what happened? They threw stuff off on the runway? What do you mean remove stuff?
This is probably for weight and balance.

I'm guessing that the pilot felt that it took more pressure than usual to lower the nose during landing, and so he wanted some of the luggage and/or cargo removed from the rear of the plane. Perhaps he was concerned that the nose might pop up while taxiing?

Maybe something shifted in the cargo bay? I don't know, this doesn't sound right to me. Once the plane is fully on the ground, I can't see why there'd suddenly be a danger of tipping up.  There was a jetblue that did it a while back, but I thought that was blamed on failing to get the rear cargo bays empty before the passengers left (and/or emptying the front cargo). I can't see how it would just tip up during taxi.

@DarthPseudonym: If the pilots noticed that derotation took an unusual amount of force (indicating a tail-heavy aircraft), they might've wanted to make sure weight was getting taken out of the tail while they still had this fresh in their minds, rather than planning on just telling the ground crew to unload from the rear first once they reached the gate (and potentially forgetting about the tail-heaviness, and consequent increased-potential-for-tippiness-while-being-unloaded, during a long taxi in).

Without knowing the specific aircraft, could there have been a rear-of-CoG fuel tank that was expected to be functional so was filled with fuel, but once in flight it was not able to be emptied (pump failed, etc.) so that fuel mass was not able to be burned off or shifted to another tank, leading to an unexpected CoG farther back than normal?

If the plane were borderline unstable on the ground, is it possible or even likely that the usual front first deboarding could lighten the plane enough for it to tip the nose up?

If you'd provide the flight number, we could look up which specific aircraft was in use and offer more insight into possible scenarios.

Thanks to all for the informative comments.  It was flight BA 1334 on 24 Sept  LHR > NCL

@SkipMiller Certainly that's possible; see the JetBlue incident I mentioned from last year.

I would say the pilot was shooting his mouth off.

@SkipMiller Yes, that is 100% a thing that can happen with the 737-900ER if the tail-support stick isn't used: https://www.avweb.com/aviation-news/update-more-details-emerge-in-united-737-900-tail-strike-episode/

There is no scenario where that would make sense. If the Center of Gravity made the airplane controllable in the air, there is no condition that could make it "tippy" on the ground taxiing in. Otherwise the airplane would be at the bottom of a hole in the ground.
My guess is the announcement was to explain why the airplane was stopping for an extended period on a taxiway, likely to have a brake inspection done following an overheat indication in the cockpit. The idea would be to avoid alarming passengers as airport vehicles approach the airplane while stopped on the taxiway.
Just a theory as to the most plausible explanation.

There's a big difference between how far back the center of mass has to be to render the plane uncontrollable and how far back it can be before causing noticeably-interesting handling characteristics, though.  That's part of why the edge of the certified envelope is well short of the point at which LoC occurs.  (Also, even if the pilots noticed that the plane was tail-heavy enough to cause noticeable handling difficulties in flight, they can't really do anything about that before landing anyway - but they can at least wait to taxi until the problem's been resolved.)

Also, wouldn't news of a weight-and-balance exceedance be more worrying (being a much-more-serious situation) than news of hot brakes?  I know I'd be much more worried to hear that the flight I was just on'd had a dangerously-far-aft CoM than to hear that the brakes were hot.

Also also, a mild excessively-aft-CoM situation might've been trimmed out during most of the flight and only become noticeable by a need for increased pitch-control inputs during landing.

@Vikki As someone not in the aviation industry, overheated brakes sounds like a much bigger problem (danger! hot! fire! explosion!) than a balance issue (who cares its not like the plane can tip over right?). I'm aware that the actual impact of an imbalance is potentially higher than some overheated brakes, but most laypeople probably aren't. To a human, getting burned is a bigger issue than tripping.

There is one aircraft for which your first paragraph is not true, the 747-100/200/400. The takeoff and landing aft c.g. limit is 33.0% MAC. However, in addition to a main landing gear at 26.6% MAC, there is a body gear aft of the main gear. If that fails to extend, flight characteristics don't change other than a little less drag, but after landing and the c.g. is greater than 26.6%, there will be a moment to set the aircraft on its tail, rotating about the main gear. As I remember, the checklist action item is DO NOT ATTEMPT TO TAXI.

@Terry: I'm slightly worried at the implication that someone somewhere at some point must've actually tried to taxi in that situation, tilted back with the tail on the runway and the nose in the air, for them to put in that explicit warning...

@Terry Well that's a gear problem, not a C of G problem.

@Vikki There is something seriously wrong with an airplane that is within its CG range yet is in danger of tipping back on the ground.  On landing, the CG will be forward of the takeoff case because of fuel being burned off, unless it lands after only the center tank has been used. Once center tank fuel is consumed, CoG starts to move forward. In any case, the gear contact point must be aft of the aft CG limit. On airplanes with tail mounted engines, it's quite a ways back, so it doesn't tip when unloaded while full of fuel, which can require a larger tail surface than otherwise.

That first para is a bit strong - airworthiness depends on the relationship between the center of mass and the center of lift. Ground handling depends on the relationship between the COM and the position of the wheels, so there is no strong link between the two.

@JohnK: "unless it lands after only the center tank has been used" - perfectly plausible if they're tankering to avoid having to refuel at a destination with really-expensive fuel.

It's quite possible for an aircraft to be tail heavy on the ground without it affecting flight performance. Usually though this is temporary during loading (so the aircraft won't fly like that, thus flight performance is not affected). That's why some aircraft have built in tail stands, and for some others temporary ones are sometimes used during loading or maintenance.

@Vikki in any case the aircraft's CofG must remain at or forward of the aft limit for the entire flight.

Many thanks for all the comments, many of which,, I confess, went over my head but I will learn.    I grilled my wife for more details.  "Did she think the plane landed faster than normal?  Yes.   'Did she think the pilot braked harder than normal?  Yes" but bear in mind she is not a regular flyer.   A couple of men rolled up but nothing was off-loaded per se.  Some minutes later, the pilot informed the cabin that he had rung ahead to get the luggage off-loaded before the passengers".

But when they got to the jetway, no luggage was off-loaded but passengers did exit by both front and rear exits.I have no idea if that information helps or hinders.

Is there any possibility that the landing itself is what caused the mass to shift? For example, if there had been something heavy in storage, and a cable failed such that it then slid backwards during the flare, and then maybe got wedged on something that prevented it from sliding forward during the braking?

@countryman Generally when an airplane has to stop on a taxiway after landing and a vehicle drives up while out on a taxiway somewhere, it's almost certainly to check out the brakes after an overheat indication in the cockpit. It's to check for signs they may be on fire or at risk of fire and you want that done while away from the terminal or other airplanes. If there was a brake fire, the capt would likely order an emergency evacuation on the spot. Could be something else, but if you made me bet $100, I'd bet on that.

@TheGuywithTheHat "who cares its not like the plane can tip over right?" Some planes absolutely can tip over if unloaded improperly: https://www.avweb.com/aviation-news/update-more-details-emerge-in-united-737-900-tail-strike-episode/

@John K.    I know all about that !!  My wife and I were flying back to LGW one time and on final approach.  I noticed that the flaps hand't extended and mentioned as much to the missus.  "Is that a problem?" she asked.  "No, just that we will land faster than normal".  A few seconds later the pilot said the very same thing over the cabin loudspeakers.   Sure enough, we came in like the clappers ...most exciting..and at the end of the runway, the pilot taxied off it and announced that 'there isn't a spare gate' and so we will be waiting here for ten minutes.

I whispered to my wife " That's not true.  He's waiting to see if the brakes catch fire".  Sure enough, a few seconds later, couple of fire tenders arrived to keep an eye out.    Ten minutes passed, we headed off the the terminal and on arrival found another fire-engine waiting there 'just in case'.    I was rather impressed by the professionalism shown at LGW that day.

@countryman You weren't in a one of the little 50 seat CRJ200s were you? Flapless landings were quite common. The reference speed at typical landing weight (the speed you fly the last part of the approach at) was over 160kts, about 184 mph.

LOL...no, full blown passenger jet.

There are too few details in the question to actually pin point the problem and eliminate extreme cases, but...

Does a passenger jet detect excessive weight in the tail

This situation can be sensed indirectly.
There is a switch in the shock absorber mechanism of the nose landing gear (sometimes called the squat switch) which closes when enough force is applied to the shock absorber, which happens when the aircraft in fully landed (a status known as "weight on wheels"). This switch can be prevented to close for different reasons, for example if there is too much weight at the rear of the aircraft, the main wheels, located under the belly, at the level of the wing, acting as a pivot.
In this case some systems may be prevented to be used and the ground proximity warning system, may issue alerts. As there is not enough weight on the nose gear, it's directional  effectiveness may be unknown, it can be be considered unsafe to use.

the pilot told the passengers that the tail was very heavy and that they needed to remove stuff

Some mass in the holds have moved rearward or there is a fuel mass imbalance in the tanks, the forward tanks are empty, but rear tanks are still full. Such fuel imbalance may have been created by a malfunctioning fuel pump in tanks aft of the main wheels.

Center fuel tanks on a A321 neo, for extended operations. Source.
In that case the crew may want to remove some mass from the aft cargo hold, or move some mass aft of the wing to a location forward.

It's hard to say what could have happened here -- assuming the pilot was being honest about what was going on, he probably felt an unusual tail-heaviness while landing and it was disturbing enough that he didn't want to go very far in that state.
How could tail-heaviness occur? There are a couple of things I can think. Based on the route you mentioned in your comment, you were probably in an Airbus A320, which (like many airliners) has a small fuel tank in the tail called a trim tank. Fuel can be shifted into the tail to adjust the aircraft's balance, but if the pump broke down, it's possible they couldn't remove fuel from the trim tank when needed, leading to an unusually heavy rear. Alternatively, it's possible something in the cargo hold shifted and they weren't comfortable with continuing until that was inspected. That shouldn't happen, but it's hard to say anything is impossible.
It seems unlikely to me that the imbalance would be enough that the airplane might tip up during taxiing, but pilots are often conservative with the safety of their plane, so if anything felt off, they would likely rather get it inspected just off the runway rather than take even a small risk of a multi-million-dollar accident. A small delay out of an abundance of caution is a small price to pay.
This may be related to the fact that just about a year ago (in October of 2023), a JetBlue airliner overbalanced and tipped up while sitting quietly at the gate, causing an expensive tail-strike:

The incident was blamed on failure to remove some heavy cargo from the rear bay before the passengers and forward cargo bays had cleared, which led to the bizarre situation. As I recall, the rear cargo door was stuck, leading to delays in unloading.

Flight history on FR24 confirms that the airplane in question was an A320.

Regardless of the status of the trim tank, a CoG far enough aft to create a risk of tipping backward on the gear after landing, the gear being aft of the CofG limit, would have been at best neutrally stable or unstable in flight. You would never certify such a design.

Yeah, I'm thinking something felt weird on landing and the pilot, possibly with JetBlue in his head, just decided it was safer to get it checked before doing anything else.


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Has a flight data recorder or cockpit voice recorder ever failed in an accident?
Has there been a time either during an aviation accident or during normal flight where the flight data recorder or the cockpit voice recorder devices failed or the data was significantly corrupted?

There are a number of incidents recorded where the FDR was destroyed or unreadable, most being due to post-accident fires.  Wiki have the following list; https://en.wikipedia.org/wiki/List_of_unrecovered_flight_recorders

@SteveMatthews-- The above comment should be made an answer.

I seem to recall a case where a flight recorder was actually stuck prior to an accident, recording nothing until a midair collision jarred it loose. So it was able to record the aftermath. I'll have to see if I can find that.

It was the Zagreb midair. I can't find anything in that article about it, but a book I have says, "The cockpit voice recorder on the Yugoslav aircraft had not been working properly but the impact jolted it into action." (Air Disasters, Stanley Stewart, p. 142)

As mentioned in the comment above, there is lot of aviation accidents with unrecovered flight recorders. List from wikipedia:
https://en.wikipedia.org/wiki/List_of_unrecovered_flight_recorders

Unrecovered is not the same thing as failed.

How do you know the recorder failed if it wasn't recovered?

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How long did the SR-71 take to get to cruising altitude?
How long did it take for the SR-71 to go from taking off to 80,000 ft, including refueling?

According to the performance data in the manual it takes 19.9 minutes to get to 70,600 feet.

Refueling time will depend on how much fuel is needed for the mission as well as how long it takes to hook up to the tanker. The manual covers the operations for that but  I dont see a flow rate of fuel in there. A lot of that stuff is covered in this podcast.

@bogl good catch, I have updated.

That's an average ROC of 3,500 fpm.  Or to put it another way, the vertical component alone of the climb is 35 knots!  That must have been like riding a tiger.

@WayneConrad, you should listed to the linked podcast, most of the backbird pilots note that aside from the high landing/approach speed the plane did not really feel that fast due to the altitude and lack of close references as you climb.I think the exact quote was "you get a greater sensation of speed at 500 ft above ground in a piper cub"

That makes sense.  Wilco, thanks.

Chapter 4 of "SR-71 Revealed" by Richard Graham states that refueling rates from both the KC-10 and KC-135Q were 6,000 lbs per minute.  He ballparks 15 minutes for a full refuel, though it seems like the operational details on refueling varied a lot per-mission, and even a single refueling rendezvous may have involved refueling from more than tanker operating in a "cell".

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How can lift be less than thrust that is less than weight?
Physics in schools teaches two contradictory and mutually exclusive things:

That the upward lift force on an airplane in flight equal its weight (Lift = Weight = Mass x Gravity). This is based on applying Newtons 2nd law of motion (F = ma) to the airplane in flight.
However, modern commercial airplanes like the Boeing 747–400 can fly with thrust-to-weight ratio as low as 0.3. Here the engine thrust is only 0.3x the weight of the airplane, but this thrust is sufficient to push the airplane forward and generate enough lift to fly.

Therefore the upward force required for lift and flight must be a lot less than 0.3x the weight of the Boeing 747-400 (Lift < Thrust < Weight). Both statements cannot be true. Which is correct?

You may read how it flies?. You are mixing 2 notions: thrust and lift. Keep in mind that thrust is not necessary to stay aloft (you can glide). Moreover you may rephrase the title of your question so that it is easier to browse through all questions speaking of physics of flight.

There are some problems with your question.  It may help you to read the related answer https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476 .  In a steep climb the 747 is surely producing much more thrust than in level flight, but in neither case does the thrust need to be anywhere near equal to weight.  Lift is LESS than weight in steady-state powered climb btw.

The root of the problem with this question is the embedded idea that lift must be less than thrust.  Hopefully some of the answers have made clear why that is not true, in most cases, and certainly in level constant-speed flight.

"With enough thrust, even a brick can fly."  The only time that the thrust-to-weight is at a 1:1 ratio with lift is when you're talking about bricks, rockets, pianos, and other things that don't have wings.

@Ghedipunk even a brick generates some lift if you throw it at the right angle.

@alephzero, very true (and have a comment uptick for pointing that out), as evidenced by the Space Shuttle.  It makes its point better than talking about spherical cows that graze on frictionless planes, though.

@Ghedipunk: Even a frictonless airplane would surely have its aerodynamics degraded by all that grass, anyways.

Your misunderstanding lies in your thought that lift is smaller than thrust, while in fact, lift is much larger than thrust.
The lift is provided by the wings. Their purpose is exactly to create a lift force (upwards force) while requiring relatively little thrust (forwards force). How well they do this is expressed by their lift-to-drag ratio (L/D ratio). A modern airplane can have an L/D of 20 or more. This means that for every newton of thrust, you can lift 20 newtons of weight!
You can think of wings the same way you think of an inclined plane: you need a lot less force to push a car up a gentle hill than up a steep hill. Indeed, if the hill is gentle, most people can push the car up the hill, while most people cannot single-handedly lift a car up. Wings work by pushing air down, and by pushing the air down at a relatively shallow angle (by going forwards real fast), you need less force to go forwards than to lift the plane.

If you want to know more than this 'intuitive' explanation, perhaps I can recommend you this answer of mine. I will present a terse derivation below, with the aim to calculate the L/D (neglection parasitic drag) purely from first principles.
Lift is due to a certain mass flow $\dot{m}$ being given a certain velocity $v$ downwards: $$L=\dot{m}v.$$ This mass flow is due to the wings encountering a certain amount of air; as an approximation, you can think that the airplane only affects a circular 'tube' of air with the diameter equal to its wingspan. The energy per unit of time (power) necessary to impart this downwards momentum equals $$P=\frac{1}{2}\dot{m}v^2.$$
This power is provided by the combination of thrust $T$ and forwards velocity, $$P=Tu.$$ This gives us an expression for the thrust: $$T=\frac{P}{u} = \frac{1}{2}\dot{m}\frac{v^2}{u}.$$ Note that the thrust $T$ equals drag force $D$, so we can now calculate the L/D:
$$\frac{L}{D}=\frac{L}{T} = \frac{\dot{m}v}{\frac{1}{2}\dot{m}\frac{v^2}{u}} = 2\frac{u}{v}. $$
In other words, like I said before, the L/D ratio is exactly how 'shallow' the relative velocity vector of the affected airflow is. It follows that the forwards speed $u$ should be as large as possible to maximize the L/D. However, I only took into account the so-called "lift induced drag" which is purely the horizontal force it takes to create a vertical lift force. In reality, there is also parasitic drag, which scales with $u^2$ and will at some point dominate over the reduction in L/D.

This is an outstanding answer in every regard.  I enjoy that it starts out with a simple analogy and then gets more serious, and finally shows how the serious analysis supports the analogy.  Also points out where the analysis breaks down.  Very nicely done.

Because a wing produces way more lift than drag.
Thrust must equal drag, not lift.

...thrust-to-weight ratio as low as 0.3.

That's a drag-to-lift ratio, actually. And that is why you find fixed wing aircraft everywhere, otherwise all aircraft would be VTOLs

Drag-to-lift ratio is around 0.05. 0.3 is about right for a thrust-to-weight at full power, which allows the aircraft to climb quite fast.

For any airplane to fly level at a constant altitude, the amount of lift generated by its wings must be equal to its weight. If lift exceeds weight, the plane starts to climb. If weight exceeds lift, the plane starts to descend. In a constant rate climb/descent, lift equals weight; a simplification that ignores the propulsive force vector.
The engine & prop in a light plane like a 4-seat Cessna develops no more than a couple hundred pounds of thrust at takeoff which is much less than the weight of the plane. But that amount of thrust is enough to move the plane's wings through the air at a speed sufficient to make enough lift to get the plane off the ground.
Note that the propulsive thrust does not have to equal the plane's weight in order to make it fly; this is only true if the plane had no wings and you were trying to hoist it off the ground by pointing the plane straight up and "hanging" it on its propeller.

Starting from level flight, if lift exceeds weight, this doesn't cause a climb but rather causes an upward curvature of the flight path-- like an entry to a loop.  Likewise weight exceeding lift doesn't cause a descent but rather causes a downward curvature of the flight path.  Lift is LESS than weight in a steady-state climb.  See related answer https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476

"upward curvature of the flight path" aka a climb? Unless a climb at increasing rate is not climby enough in your Book of Climbs.

@AEhere -- sure it's plenty climby but it cannot be sustained for more than a few seconds -- you will either run out of steam as you approach vertical or you will go around past inverted in which case you are no longer climbing!

"If lift exceeds weight, the plane starts to climb. If weight exceeds lift, the plane starts to descend." This is not accurate. Lift (or, well, total upward force, part of which could come directly from thrust) = weight is true whenever the aircraft is not accelerating vertically. A climb at constant vertical speed will have upward force = weight, as will a descent at constant vertical speed. Slowing a decent rate will have upward force > weight, even though the aircraft is still descending. Slowing a climb rate will have upward force < weight, even though the aircraft is still climbing.

When you push your car on a flat area you need very little force to move it, much less than its weight
With the same force you are pushing the car, you notice that at the beginning it will get an accelerated motion and afterwards due to the drag mainly, the speed becomes steady even though keeping the same pushing effort
Therefore in order to make an aircraft accelerate you need just to overcome the drag
Acceleration means increasing the speed, thus the drag, but all you need is to overcome the drag not the weight.
To lift off you just need to get the necessary speed for the wings to provide, as a function of the angle of attack, that lift that is just above the aircraft weight

I agree with this answer, except for the statement of not having to overcome weight. A car can sit on a flat surface, but the aircraft must stay at an incline to remain afloat. This incline may be small (like for a glider) but it means the plane must have some amount of energy expended at all times to fly. The amount of energy used is strongly linked to weight, in fact it's a squared component in the lift/drag equation. So I think that could be reworded, otherwise great answer!

Not a disagreement but rather a clarification that you may wish to include in your answer.  Yes lift may slightly exceed lift at takeoff but only because the flight path is CURVING upwards, i.e. bending up into a climb.  A curvature is a from of acceleration and requires force to accomplish, i.e. an excess of lift over weight.  Once the climb is established and stabilized, lift is actually a bit less than weight.  See related answer https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476

@YAHsaves, thank you, my answer if for beginners as you may imagine from the question content, so I had to explain the matter as intuitively as possible.

@quiet flyer, thank you, my answer if for beginners as you may imagine from the question content, so I had to explain the matter as intuitively as possible


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Term to differentiate between a flight and a flight instance?
What's a term to differentiate between a flight schedule (QF 72) and a particular instance of that flight schedule (QF 72 departing on July 25)?
For example, I might ask "Which flight are you taking", and you might reply "QF 72". What I really meant was "Which scheduled instance of QF 72 are you taking?"
The flight schedule and a specific instance of that flight schedule are both referred to as a "flight". What are the terms that more precisely differentiate between these two concepts?

You might say "the QF 72 service between Singapore and Perth" for the flight schedule, and "the QF 72 flight on July 25" for a particular instance of the flight.
The term service used in this way is seen more often in other transportation sectors such as bus and train. But it can also apply to scheduled air services.


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Private plane without a flight number - is it illegal?
This article from the Detroit News describes the arrest of a rapper on various charges, which aren't relevant here.  However, there are a couple of aviation-related statements in it that seem to imply additional offences were committed:

Agents soon discovered the private jet did not have an assigned flight number.

Agents asked airport security to confirm [the accused] was aboard the flight. They checked the manifest but [the accused] was not listed among the passengers.

"He has access to a private plane with a pilot willing to fly an unnumbered flight where [the accused] is not included on the flight’s manifest,” [the detective] said. “That’s not an accident. That’s not a happenstance.”

Has the pilot (or anyone else) actually committed an offence in these circumstances?  If so, which particular provisions of the FAA regulations have been violated?  Or is it just the newspaper (or the police) introducing irrelevant material to make the accused's conduct seem worse than it actually is?

Flight number. Flights operated in the United States are not required to operate using an airline callsign and flight number. All aircraft are required to have a registration number, similar to a license plate. Flights may use a flight number, which could be assigned to any aircraft, or may identify themselves using the aircraft registration number, which is unique to the aircraft. There is no legal concern here.

Passenger manifest. To protect the flying public, the FAA regulates operators who carry passengers for hire. There are several different sets of regulations which may apply to an operation. More information can be found elsewhere online, for example this article from Pilot Institute. However, as a brief overview:

14 CFR Part 121 applies to scheduled air carrier operations—you would recognize operations carried out by "mainline" airlines such as American and "low-cost" airlines like Sprit, as well as "regional" airlines like Endeavor or Republic. These regulations are most restrictive.
14 CFR Part 135 applies to "commuter operations and on-demand operations," commonly known as "charters." These would generally be smaller twin-turbojet aircraft which are operated for the benefit of anyone who buys a flight from Point A to Point B; Executive Jet Management is operated under Part 135. Some airlines, for example Cape Air, also operate smaller turbo-propeller flights under Part 135.
14 CFR Part 91K applies to "fractional ownership" flights. This is a system in which customers technically purchase part ownership in an aircraft or fleet of aircraft, and therefore have the right to (additionally) purchase a certain amount of flight time per year. NetJets operates aircraft under Part 91K. These regulations are less restrictive.

A "private aircraft" as described in the article would likely be operated under Part 91K or Part 135.
For Part 91K, 14 CFR § 91.1027 - Recordkeeping applies:

(c) Each program manager is responsible for the preparation and accuracy of a load manifest in duplicate containing information concerning the loading of the aircraft. The manifest must be prepared before each takeoff and must include -

The number of passengers;
The total weight of the loaded aircraft;
The maximum allowable takeoff weight for that flight;
The center of gravity limits;
The center of gravity of the loaded aircraft, except that the actual center of gravity need not be computed if the aircraft is loaded according to a loading schedule or other approved method that ensures that the center of gravity of the loaded aircraft is within approved limits. In those cases, an entry must be made on the manifest indicating that the center of gravity is within limits according to a loading schedule or other approved method;
The registration number of the aircraft or flight number;
The origin and destination; and
Identification of crewmembers and their crew position assignments.

You will note that identification of the passengers is not required; however enumeration of the passengers is.
For Part 135 operators, 14 CFR § 135.63 - Recordkeeping requirements applies; the requirements are exactly the same as 91.1027 above.
Regardless of whether the flight was operated under Part 91K or Part 135, the person in question should have been accounted for on the aircraft load manifest. However in neither case was he required to be named on the manifest.

Thanks!  From the rest of the article, the flight appears to have been a perfectly legitimate Part 135 operation, with no sinister overtones.


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What is the procedure following a near miss?
This is a follow-up to this question - in particular, to this video posted in MichaelK's answer.
What would be the correct procedure for the pilot to follow in this situation?  (That is, after a near miss).  Should they report the incident to ATC immediately, even though the incident is over?  If there are differences in the procedure depending on jurisdiction, what would it be in the USA and the UK?


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What phenomenon causes the increase of drag with rain?
I know that rain generally worsens the aerodynamic propeties of an aircraft.
I'm wondering what processes cause the increase in drag.
I'm mostly interested in General Aviation, thus relatively low speed and low altitudes.
Several things I can imagine, although the effects might be small:

The water in the air makes it heavier, increasing the effective value of the air density, $\rho$
Rain might lower the temperature, again having an effect on $\rho$
Rain drops on the wing might cause earlier transition on the wing, increasing  drag.

I hope somebody can clarify what magnitude these effects have, and if there are any more effects that cause an increase in drag when flying in the rain.

On the glider rain drops can reduce glide ratio substantially. It depends on the wing profile but it can drop from 45 to 30.

@Andrius, I know rain can change the glide ratio significantly, but I'm mostly interested in what causes these changes. I'll reword my question to make it a bit clearer. I'm also focusing more on the increase in drag, to keep a narrow focus.

For us physics challenged readers, what does $\rho$ stand for?

Sorry, $\rho$ is the density of the air

While Peter Kämpf already commented the effects I'll take a look at the numbers. Don't expect this calculations to be exact predictions, rather worst cases to know the magnitude of following effects:

Mass increase due to rain drops
Forces due to drops hitting the wing
Early tripping of the boundary layer

After deducing some equations I'll feed them with the data of two different airplanes.
Mass increase due to rain drops
The dew layer you find on your airplane in the morning can reach a maximum altitude of 0.8 mm. Raindrops, which do not fill the complete wing surface, don't usually exceed 2 mm diameter.Therefore taking an average distributed height of $h_W=1mm$ for the water on the wing should be still a worst-case prediction.
$$m_{Drops}=\rho_{W} \cdot A\cdot h_W$$
$A$ is the projected area of the aircraft on the horizontal plane and $\rho_W$ water density.
Forces due to drops hitting the wing
Normal rainfall intensities are about 5 mm/hour and we already speak about violent rain with precipitation rates beyond 50 mm/hour. But let us take the heaviest rate ever measured: $I_R=38mm/min$.
Drops fall normally with $v_{Drop}=10m/s$.
To simplify the calculation I'll assume vertical rain and that once a drop hits the airplane it sticks to it. In reality you don't accelerate every drop to the airplane speed nor slow its vertical speed down to 0.
i) Drag increase
$$ \Delta D= \rho_{W} \cdot I_R \cdot A \cdot v_{airplane}$$
ii) Lift decrease
$$ \Delta L= \rho_{W} \cdot I_R \cdot A \cdot v_{Drop}$$
To calculate the total drag ($D=C_D\cdot \frac{\rho_{Air}}{2} V^2 \cdot A$) I've used $C_D=0.035$. $V$ is the cruise speed of the aircraft.
Early tripping of the boundary layer
While it's possible to find airfoil polars in rain searching the Internet, I'll first look at a "bugs influence" diagram where you can see the effect of small disturbances at the nose without other effects overlayed. Bugs and raindrops are similar in size, therefore the comparison should be valid.

Source
The polar above shows how the drag can double (100% increase)just because of some small disturbances at the leading edge. Keep in mind that in modern profiles the effect will probably be smaller. The next polar is from a transport aircraft airfoil.

Source
While not so important as in the glider airfoil, the drag coefficient increase (experimental results) is everywhere at least a two-digit percentage.
Airplanes
I've taken two airplanes with really different wing loadings: A1 would be something between the two Solar Impulse models and A2 a single engine four-seater like a Cessna 172. The last three rows give the drag, lift and mass difference as described above as a percentage of its total value.
|                       |            A1 |     A2 |
|-----------------------+---------------+--------|
| Mass m [kg]           |          1600 |   1000 |
| Area A [m^2]          |           210 |     20 |
| Wing loading [kg/m^2] |           7.6 |     50 |
| Cruise speed V [m/s]  |            20 |     58 |
| Lift L [N]            |         15696 |   9810 |
| Drag D [N]            |          1800 |   1323 |
| Drops mass mW [kg]    |           210 |     20 |
| Drag increase dD [N]  |          2660 |    703 |
| Lift decrease dL [N]  |          1330 |    127 |
|-----------------------+---------------+--------|
| dD/D [%]              |           147 |     53 |
| dL/L [%]              |           8.5 |    1.3 |
| mW/m [%]              |            13 |      2 |
|-----------------------+---------------+--------|

SUMMARY
For a normal GA airplane the water mass on the wings and the lift decrease influence due to raindrops hitting the wing even considering the worst possible conditions are low single digit percentages.
While drag increase due to drops hitting the wing is relevant in the extreme case I took here, under normal conditions it should be below single digit percentages too.
On the other hand the drag increase due to the early tripping of the boundary layer is at least a two digit percentage, at high lift coefficients even more.
Airplanes with lower wing loading are more affected with rain precipitation.

Fly a Schempp-Hirth Janus or a PIK-20 into rain. The effect will be dramatic. Make sure you have some place to land nearby! In both gliders your rate of sink will triple and the minimum speed will increase by maybe 20 km/h, so make sure you come in fast for landing while the wing is still wet.
The reason is the airfoil, the Wortmann FX 67-170. With 17% relative thickness and a very long laminar boundary layer, the recompression cannot be handled by a boundary layer which has been tripped near the leading edge. The consequence is flow separation already at moderate angles of attack if the wing is dirty from bugs or rain. The flight performance changes dramatically.
Then perform the same with a Schempp-Hirth Discus or an ASW-20. Sink speed will increase, but the effect is much less dramatic and the minimum speed will barely go up.
There is no general rule of thumb, and two of the three reasons you suspect are indeed the major effects: Added mass from the water film over all upper surfaces and early tripping of the boundary layer. In big aircraft where the transition happens close to the leading edge anyway the difference made by rain is very small. It might even help by increasing engine thrust: Rain will cool the air and by absorbing heat when vaporizing will lower the temperature rise in the compressor of a jet engine. Note that many early jets used water injection to increase engine thrust on hot and high airfields.
For small aircraft with their lower volume to surface ratios the mass increase is more noticeable, and if the airfoil was optimized for minimum drag in clean conditions the early transition will effect a dramatic change in a few cases. The airfoils of most powered aircraft are not nearly as sensitive to rain. Normally, those aircraft are flown VFR, and the drop in visibility caused by rain will be the biggest effect.

This is a very good answer to the question specifically asked, but I think that there is an implied, more broad question about how rain effects "performance", rather than just increased drag.
What I am referring to, is the obvious fact that rain is "falling". There is a force applied to the aircraft in a downward direction by both the rain, and the accompanying downdraft.

@GregTaylor I tried to keep the question specific, which is usually good for the quality of the answers given. If you want to ask about the forces of rain falling on the aircraft, you're free to do so. When doing some research on this question I found answers to such a question here:
http://www.airliners.net/aviation-forums/tech_ops/read.main/317372/
This answer suggests that any forces caused by the rain are very small

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Where is the APU in the case of an S-duct (like the L-1011)?
This question asks about the S-duct of the L-1011 Tristar. Fooot's answer shows a cut-out of the L-1011. Looking at this cut-out, I'm wondering:
If the engine is at the 'normal' location of the APU, where did the designers of the L-1011 put the APU?
I did a search, and found this site which mentions

The APU is located in a compartment in the aft fuselage.

Another site mentions

Its still in the "tail". Its mounted below the #2 engine. Just forward of the tail skid.

But I don't see how you could fit the APU in the small available space.

(Source via aircraft.wikia.com)
APU is number 134 on the cutaway, which you can find near root of left stabilator leading edge. It is ahead of the engine, below the intake duct, on the left side.
By Reedy (Own work) [CC BY-SA 3.0], via Wikimedia Commons
Its exhaust is on the right side close to the stabilator's leading edge.

The APU on the B727 was placed inside the main gear wheel well.

image source
It drew air from inside the wheel well, so it was only for use on the ground. The exhaust was on the top of the right wing close to the fuselage.

image source

And man is it loud. SOURCE: worked on them for 20 years.

@SportsRacer Yes they were. Used to work at the FedEx hub back when the fleet was mostly 727's. The head cans we wore were not enough working at the belly cargo doors.

Does this mean the APU  in the 727 couldn't be used to provide emergency power in air?

@TomMcW: Did they not have ground power in Memphis?

Some places on the ramp probably did. When I was there a large portion of the 727's were on remote stands.

@zymhan Just saw your question. The APU on a 727 cannot be operated in flight. Ground use only.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the first fighter jet which was built with twin engine and tail configuration?
What is the very first fighter jet that pioneers the twin engine and twin vertical stabilizers over the engine just before the exhaust (like the F-14, F-15, F-22, Su-35, etc)?

as from the current answers can be seen, you might want to specify whether you are looking for also prototypes or just fighters that entered service?

Both are okay, but don't forget to clarify which one was a prototype or service aircrafts.

The Heinkel He-280 was a twin-engine twin-tail jet fighter, but like most early jets had podded engines

Not a fighter, but twin-engine and tails-over-engine: SNCAC 1071, French/1948.

One of the first was probably the unconventional XP-79.
https://en.wikipedia.org/wiki/Northrop_XP-79
It never entered service of course, but does meet your criteria.
Something that kinda meets your criteria would be the F7U Cutlass. Twin tails and twin rear mounted engines though the tails were in separate tail booms outboard of the engines, a similar configuration to that of the DH Sea Vixen and Venom of roughly the same period.
These did enter service.
Then there is the XF-90 which has twin rear mounted engines with a single vertical stabiliser over them.
The Soviet La250 and Tu28 had similar configuration but came later.
This also did not enter service.

Probably the MiG-25.  Near as I know, it was the first to use the ‘four poster’ layout.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

@Sean Of course it answers the question. It might not be a good answer without listing any source for this, but it still is an answer.

Downvoted because of the probably. Since you probably meant to revise this later, I look forward to upvoting once this is a solid answer. Probably.

Prototype only: the YF-12A. This was an A-12 derived fighter plane. The A-12 is better known for its other derivative, the SR-71 Blackbird. Both the A-12 and the SR-71 were intended for reconnaissance tasks,  but the YF-12 would have been a fighter.
With a first flight in 1963, it beats the Mig-25 by a year.

Possibly might be the WW-2 era Me-262?  I believe the prototype had twin tails.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

@Sean I don't think you're using the "not an answer"-review feature quite right. It's meant for posts that don't come close to being an answer, not for posts that do not meet your criteria for being a good answer. For that you can leave a normal comment, or even downvote.

I would agree that this is more of a comment than an answer.

@Sean it is an answer. It's not the correct answer, but it is an answer.


text:
Why are wind tunnel models sometimes placed upside down?
Sometimes in wind tunnel tests, models are suspended upside-down:

Image obtained from DNW - Interference corrections

Image obtained from this Youtube video of testing at AEDC

Image obtained from UWAL - Model Support Effects
And sometimes they are in the normal flying position:

Image obtained from TRI models - wind tunnel testing
What is the specific reason for putting wind tunnel models upside-down?
I thought it had something to do with minimizing the influence of the support, but putting the plane upside down doesn't change much about the location of the support.

It looks like there's a Storm Trooper in that second picture!

@FreeMan it looks more like a screaming frog with really long arms :)

Its because to test to see if the plane could fly upside down. It could happen so they need to test just in case. Test inspired by Will Smith In the movie FLIGHT.

In the times before strain gauges and electronic transducers, the forces on the wind tunnel model were measured with scales. To avoid interference as much as possible, the model was hung on strings, the scales being mounted above the measurement area of the tunnel. Since the forces in upright flight were of the highest interest, gravity plus lift would make sure that the strings were always taut. If the model were hung upright, the lift might grow bigger than the weight, and the model would move towards the scales.
Today, the orientation of the model can be either way, and only in exceptional cases there is a strong reason for placing it upside down. This might be when unusual attitudes together with the supporting sting (the long spike on which the model sits) would block one part of the wind tunnel cross section too much, or if movement restrictions of the sting force a certain orientation.
Another reason is flow visualization techniques: Sometimes it helps to put a coat of a mixture of petroleum and pigments on the surface and to watch where the pigments are blown by the flow (this is called an Anstrichbild). Obviously, it helps to place the surface of interest on top.

But in the first picture from the DNW the model is fixed using a rigid support, so any movement is prohibited. Yet, still the model is fixed upside down.

@ROIMaison, that could still be a power of habit.

@ROIMaison: Those aren't really rigid supports. If they were rigid they'd defeat the purpose of the wind tunnel. They're strain gauges which is a springy connection that measures forces (exactly the same as a digital kitchen/bathroom scale but more expensive). Depending on the design of the gauge/support, you may need to install the plane upside down either for the same reason as with older mechanical gauges or because one direction have better resolution or noise cancellation than the other.

@slebetman: I understand that they're not actually rigid, but I used the word rigid to convey the contrast with wires.

@ROIMaison: My point is the contrast may not be that big. Springs are springs be they thin coiled wire springs or straing gauges. It depends on how you design the spring.

@slebetman: Try to look at the readout of strain gauges which sit under 45° on a solid metal rod, then torque the rod with your bare hands. You will not see any torsion, but the strain gauges will very clearly indicate that something is happening. They are extremely sensitive, and the sting of a wind tunnel is just such a (hollow) steel rod. For all practical purposes it may very well be called rigid, even though its strain gauges will give good readings.

The main reason for placing the wind tunnel models upside down was that it was easier to measure lift that way.
The first wind tunnels faced two main problems while measuring body forces:

How to avoid interference- This was overcome by either suspending the model on wires or placing the model on a metal pole called stinger that projected to the rear side of the model.
How to measure the forces accurately.

The forces on the initial wind tunnels were mostly measured using force balance. In case of the models suspended on wires, 'tare' weight can then be set and any additional balance weight required is the lift.

Though the wind tunnels became more sophisticated,the models were mounted upside down both as force of habit and as measuring equipments were designed for that.
Of course, the lift could be measured the other way too and some wind tunnels followed that. There is no fundamental difference between the two. Certain wind tunnel tests like separation studies require the models be mounted in a particular (usually right way up) orientation.
In today's wind tunnels using strain gages among other things, the models can be mounted in any required orientation.

"Certain wind tunnel tests like separation studies require the models be mounted in a particular (usually right way up) orientation." I'm guessing this is because of the sensitivity of separation? Does this mean that gravity has a small but significant influence?

The mounting of models for (flow) separation studies is determined by the visualization technique used. In the scales seen in aviation, gravity is not a significant factor, though it is in fields like geophysics. On the other hand, it is important to take into account the effect of gravity for payload separation studies.

This answer is a good one, but the final paragraph is misleading for two reasons : 1) many of today's wind tunnels were built decades ago, and 2) the direction/orientation can still be important because of the resolution or range of the tunnel sensors.

You are right, especially the first part. I should've qualified the statement.


text:
Why the ground effect has such a different outcome for wings and cars?
I have some confusion concerning the ground effect.
If we take two applications: Aircraft and Cars, they seem to have completely opposite effects.
Aircraft
From Ground effect (aerodynamics), we can read:

During normal flight, the upper wing surface experiences reduced static air pressure and the lower surface comparatively higher static air pressure. Flying close to a surface increases air pressure on the lower wing surface, nicknamed the "ram" or "cushion" effect, and thereby improves the aircraft lift-to-drag ratio. The lower/nearer the wing is vs. the ground, the more pronounced the ground effect becomes.

I can understand this, the air gets 'trapped' between the wing and the surface (and as consequence the pressure on the underside of the wing goes up)
Cars
However, from Ground effect (cars) we read:

Designers shifted their efforts at understanding air flow around the perimeter, body skirts, and undersides of the vehicle to increase downforce with less drag than compared to using a wing.
This kind of ground effect is easily illustrated by taking a tarpaulin out on a windy day and holding it close to the ground: it can be observed that when close enough to the ground the tarp will be drawn towards the ground. This is due to Bernoulli's principle; as the tarp gets closer to the ground, the cross sectional area available for the air passing between it and the ground shrinks. This causes the air to accelerate and as a result pressure under the tarp drops while the pressure on top is unaffected, and together this results in a net downward force. The same principles apply to cars.

The cross sectional area decreases, causing a higher velocity. This higher velocity decreases the pressure, causing the tarpualin to go down.
This confuses me, as it seems that the same situation (a plate in the vicinity of a surface) causes two completely opposite effects.
Both the "trapping air -> pressure goes up"  as well as "trapping air -> velocity goes up -> pressure goes down" explanation seem reasonable to me.
This made me wonder:
Why the ground effect has such a different outcome for wings and cars?
If I have a plate under an angle close to the ground, will the air be trapped, creating an air cushion pushing the plate up, or will the air accelerate, causing the plate to be pushed down? Does it have to do with some degree of area decreasing?

"How come" is lousy english writing, please change to "why" on all occurrences.

@quietflyer thanks for the comment. Why is it considered bad english?
It seems to be quite common, see this Google Ngram

@quietflyer I don't see why informal writing should be discouraged on Av.SE. There's no reason to be so pedantic (and if there is, do have the courtesy to use proper grammar and capitalization).

It actually has exactly the same outcome! Well, in reverse.
Wings generate lift by forcing air down (one way or another, let's not go into details). When the wings are close to the ground, the air that is forced down can't get out of the way, resulting in a higher pressure.
Cars generate downforce by forcing air up. When air close to the ground is forced up, it can't be replaced by 'new' air easily. This results in a low pressure, sucking the car down.
The ground effect is thus exactly the same in either case: air that is either pushed out or pulled in, cannot be replenished quickly enough due to the ground being in the way. The difference in outcome is because a wing is attached at a positive angle w.r.t the ground, whereas a car is shaped like a very thick wing at a negative angle.
A comparison could be a hovercraft on the one hand, which is lifted off the ground almost exclusively by ground effect (if you were to drop a hovercraft from a plane, it wouldn't matter much whether it was powered up or not). On the other hand, a vacuum cleaner, which is sucked down to the floor exclusively by ground effect (indeed, Feynman's sprinkler experiment will tell you it is not sucked down because of the upwards acceleration of air - only the 'engine' part of your vacuum is forced down by momentum exchange. Test this by holding the vacuum in the air - you won't feel it 'pull' at all).

Then I have another related question. Let's assume we have a flat plate at the bottom of our car. However, using panels at the side (shielding the tires for example) we create a narrowing channel. This will cause the velocity to rise in the throat, and as an effect the pressure drops, and more downforce is generated. However, no air is forced up or down. If downforce is only generated by moving air up, how come only an increase in velocity will also increase the downforce?

Sorry I confused you; I meant to explain that the ground effect is not due to momentum exchange but due to pressure increase or decrease. How exactly this pressure is decreased is not addressed; I used forcing air up as an example to stay close to the airplane analogy.

Wings generate lift by being forced up by air, not forcing air down.

@ptgflyer that's Newton's third law for you.

They don't force air down unless the wing is tilted up. They form an area of low pressure, which makes the high pressure area push up the wing. What is pushed down? (I'm not referring to upwards movement being hindered.)

@ptgflyer I was hinting at the concept of downwash. Newton's 3rd law says that, for the wing to be pushed up, something has to be pushed down - compare the ground denting slightly when you stand on it.

Oh, okay. more words to fullfill length requirement

Because they are designed differently. As you've yourself pointed out, while an aircraft produces lift, the racecars produces downforce i.e. lift directed downwards. The following figure shows the design of the wing of an aircraft and the front wing of a racing car.

Image from Ground Effect Aerodynamics by Erjie Cui and Xin Zhang
Basically, the downforce of the cars is the lift towards the bottom.

When the wings are close to the ground , two things happen- the effective span is increased due to the interference of ground with vortices (within a wingspan).

Source: ghdstudios.com

Still closer to the ground (around 1/4th span), the air flow between the wing and the ground is compressed to form an air cushion. The pressure on the lower surface of the wing is increased creating additional lift.

The net result is the increase in lift and decrease in (induced) drag.
In case of cars, the same thing happens, with the force in the downward direction. As the air is 'trapped' in the the gap between the suction surface (underside in case of a car) and the ground, it initially accelerates. As the height is reduced, the flow accelerates more, resulting in less pressure and more downforce.

I think there's more to it than simply turning the wing upside down (as you do in your first picture). I agree with the wiki when it says:
"Starting in the mid-1960s, 'wings' were routinely used in the design of race cars to increase downforce (this is not a type of ground effect)"
At end you mentioned that close to the ground the air flow between the wing and the ground is compressed causing an air cushion, why wouldn't this happen at the underside of the car?

@aeroalias What about normal cars that don't have wings that produce negative lift? They also experience a ground effect.

@mezzanaccio The down-force can be produced by the body of the car. In fact, around half of the downforce in a racing car is produced by the body. Wings are used because they are simply more efficient.

@ROIMaison Because wings create a downwash, which is not done by the cars. This basically creates something akin to ram pressure in a wing near the ground, which is absent in the car, which has a flat surface.

Remember that lift means the downward acceleration of an air mass. This bends the aerodynamic force vector backwards, so a small part of the pressure force acts against the direction of motion.
If the ground blocks this bending, less of the aerodynamic force will be felt as drag. This is true regardless of the direction of this force - it can be up or down. The closer the distance between wing and ground is, the more this effect can be observed.
The difference between the ground effect of an aircraft and that of a car is their angle of attack and the resulting direction of lift: Up in case of the aircraft and down in case of the car. The magnitude of the resulting force is not changed much, only its backward tilt is reduced, which results in a dramatic drag decrease.
The car body acts also as a wing, albeit one with a small aspect ratio. All the racecar designers did is to replace the top-mounted wing (which did not feel much ground effect) by the car body itself (which is much closer to the ground). By keeping the entry area under the front bumper smaller than the exit area under the rear bumper, the car body acts like a wing at a negative angle of attack, and ground effect makes its downforce creation quite efficient.

Widening gap between car body and ground in side view (picture source). Skirts make this more efficient by preventing air entering from the side, but in the ever evolving tug-of-war between race car engineers and regulators they and some aspects of car bottom shaping have become less efficient in recent times.

But if I follow this explanation, then the example with the tarpaulin does not make a lot of sense to me... If it moves to the ground, does it mean that it accelerates the air upwards?

Then your understanding of the tarpaulin experiment is not correct. It is a bad example, anyway, because it is floppy while wing and car are stiff. The idea is to hold the windward side of the tarpaulin close to the ground, and then the wind will bend it down. If you move the whole tarpaulin down on all four edges symmetrically, no such predominant motion is visible; instead, it will flap randomly.

I hope you can help me improve my understanding. In the commens below this answerhttp://aviation.stackexchange.com/a/22643/4197, I give an example of a flow that has no vertical momentum exchange, yet provides an upward force. This gives me some issues with the air deflection explanation..

I think I understand it now.
In the case of the wing, there is a momentum exchange, and this effect is made stronger by the ground effect.
In the case of the case, at the bottom plate,(if we assume a straight bottom plate) there is no momentum exchange, but due to the narrowing channel under the car, the flow accelerates, causing downforce. Correct?

@ROIMaison: The momentum exchange is only part of the process - in the end, the momentum of the air causes a pressure field on the ground. In ground effect, the same happens, but much less air and less motion is involved, because the ground is so much nearer. Your example doesn't create momentum-free lift, just lift which involves less motion of the air.

One thing still puzzles me. You mention lift is the result of downward acceleration of mass. Then, if the ground is close, and blocks the downward flow, wouldn't we get less downward acceleration and hence less lift?

@ROIMaison: Air is still accelerated downwards, but the resulting movement is stopped by the ground before the usual trailing vortex system can develop. This is not much different from free flight, only the deceleration happens next to the wing and over a much smaller ground area.

The main difference between the two cases is that the wing generates lift by inducing a downward wind component. When this downward wind component is close enough to the ground, an increase in lift will occur according to the explanation you posted.
The two effects have the same name, but are explained by different physical principles. Just think of the distances between a wing and the ground (a few meters) and between a car chassis and the ground (a few centimetres). A car is not a lifting body and will not induce a downwind, therefore this will not cause an upward force.
The car ground effect is just "another" ground effect, which is explained correctly by your quote.

It depends on the design of course, but a car actually is a lifting body.  The whole purpose of wings, spoilers, and front air dams is to create downforce to counteract the lift produced by the body at the higher speeds of a race car.  I got my '83 Mustang GT up to 120 once and could actually feel the car rising and getting lighter on its tires.  Not something you want in the corners!

There have been some great answers here with awesome information, but I don't think any of them address the premise of the question, which I believe to be incorrect.
The premise is that it is ground effect that is creating a force keeping the car on the ground. This is not accurate. A race car sticks to the ground because it is designed to overcome ground effect.
An airplane in ground effect can still easily be flown into the ground with a sufficiently negative angle of attack, or from lack of lift due to wing stall. A car can be made to stick to the ground with suffient downforce - despite ground effect, not because of it.
If a sufficient flow of air is allowed under the car, with a positive angle of attack, it will absolutely experience the same exact ground effect as an airplane does.
Conversely, an airplane can be driven all the way down the runway at cruise speeds without lifting off if you hold the elevator all the way in, though I wouldn't recommend doing that. Ground effect is not a magic levitation force, it simply reduces drag and it can easily be overcome. If not, landing would be impossible.
The bodywork of a race car is designed to overcome ground effect. The bottom of many race cars are also designed to maximize the advantage of low pressure, but this is not ground effect. This effect can be described anecdotally with the child's experiment of placing newspaper flat on a table on top a ruler, with the end of the ruler off the table, and then striking down on the ruler, which will break the ruler thanks to the air pressure on the top of the paper. When the car tries to lift off the road, the same force helps keep it stuck - but that force is not the "ground effect" that we refer to when flying an airplane close to the ground.

+2 if I could. This is the only correct answer to the question, IMHO.

@HannoBinder: Why? Ground effect is mostly about reduced induced drag, not about lift. The direction of lift is secondary. Greg's and your understanding of what ground effect is is different from everybody's else, IMHO.

Odd, that you would think that since I even said the same about reduced drag that you just did.

The more I've thought about this the more I think the problem is that ground effect is an overloaded term. Car guy are describing a totally different force when they use the term ground effect.

@PeterKämpf I'm pretty sure it's about both; Wikipedia says "improves the aircraft lift-to-drag ratio", reason: less vortexes and more air cushion under the wings, and "While in the ground effect, the wing requires a lower angle of attack to produce the same amount of lift."

"ground effect can give the pilot the feeling that the aircraft is "floating"" seems to support the relevance of increased lift. Reduced drag would only be perceived as longitudinal acceleration, wouldn't it?

@HannoBinder: No, reduced drag is perceived as the aircraft floating down the short runway and not settling down in time. In this situation half a degree angle of attack less is irrelevant, but the fact that I am about to overrun and crash is very much more relevant.

I think there is a lot of great information in the answers here but they miss the obvious. It's about the angle of attack.
Even a wing angled to point downwards close to the ground will not generate lift - it will create a lower pressure under/behind the wing (as well as a greater pressure above) and "suck" (and push) it down into the ground. And that is what is desired and designed with cars.


text:
What is the one-engine-inoperative maximum range flight configuration?
I was talking with a friend about what things would happen if you experienced engine problems in cruise. We were wondering the following:
What things would need to be changed to arrive in the optimal configuration for range?
Here I found that the speed is reduced to approximately 80% of the normal cruise value, but I can image that more things change, perhaps the altitude?

The optimization problem is unchanged from the all engines operative (AEO) cruise optimization. You should still:

Climb as high as practical (colder, thinner air is better for range)
Set the lift coefficient such that drag is minimized for propeller aircraft ($c_{Di} = c_{D0}$). For turbojets, minimize energy per flight time ($c_{Di} = \frac{1}{3}\cdot c_{D0}$). For turbofans, pick a value between those two extremes.

For the computation of the theoretical optimum polar point please see this answer for a complete derivation including wind, or this answer for a shorter explanation.
Your flight altitude will certainly be lower, because you need to generate more thrust on the remaining engine (requiring higher density), which in turn will result in a lower TAS (true air speed). The CAS (calibrated airspeed) should stay about the same, however. The configuration (flaps, gear) should stay unchanged, only rudder deflection will be added for asymmetric thrust compensation. The resulting sideslip will increase the zero-lift drag of the airplane a little, so a slightly lower speed than in the AEO case would result.
In the unlikely case that the required density will result in an altitude below ground level, you will need to slow down and need to set flaps to stay airborne, and now the new optimum with flaps will result in a markedly higher lift coefficient because deploying flaps will increase the zero-lift drag quite a bit. If you have the flaps-down polar, the optimum is easy to find.

Your guess is right.
In this article you can read about engine-out procedures.
Because you can't maintain a certain altitude without sufficient thrust, losing an engine means you can't maintain the same cruising altitude without all the engines running.
This is, of course, based on the fact that you're flying around the maximum altitude you can fly with both the engines operative given a certain aircraft weight.
It's worth nothing that when losing an engine on modern twin-engines airliners, usually the APU is started to provide electric power and pneumatic pressure (also, hydraulic pressure via electric pumps that runs on electric power produced by the only remaining engine's generator), but usually APUs can't be started and/or run over a certain altitude (20.000 - 25.000 ft, depending from the aircraft), so descending to lower altitudes is somewhat mandatory.

Do you know at what point the engine operates? Is cruise thrust the maximum thrust and as such, losing an engine will always result in 50% loss of power, or is it possible to squeeze more newtons out of an engine, but at the cost of efficiency?

Usually there are various power settings for the engines: TOGA (Take-Off/Go-Around) is the highest one, and it' used rarely (as it stresses the engines a lot). MCT (Maximum Continuous Thrust) is an high thrust power setting but can be maintained for longer timespans than TOGA. MCL (Maximum CLimb thrust) is the highest thrust setting used during climbs.
Every aircraft has it's own thrust settings (and relative names, on Airbuses for instance, there is also the FLEX power setting) and it's own engine out procedures, that usually involve using an higher thrust setting on the remaining engine.


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What is the aerodynamic influence of prop spacing and placement in multi-prop engine setups
I saw this question Is there a rule of thumb for spacing between jet engines?, which referred to How do you choose engine placement for under wings engines?, but I was wondering how the engine spacing is approached for propeller aircraft.
There are some examples as shown in this question (How far should the two propellers on a twin-engine plane be apart?), but I'm looking for a more systematic approach.
Compared to jet engines, propeller engines are larger, and have a much larger stream tube going into the engines. This will probably make their performance much more sensitive to the lateral gap between the engines.
I found this paper Aerodynamic interaction between propellers of a distributed-propulsion system in forward flight, which indicates for a three propeller setup:

At the thrust setting corresponding to maximum efficiency, the efficiency of the middle propeller is found to drop by 1.5% due to the interaction with the adjacent propellers, for a tip clearance equal to 4% of the propeller radius.

This is one data point, but I wonder how the influence varies with prop spacing.
It makes sense to me that at a far-enough spacing, the engine can be treated as independent. I'm wondering what this far-enough spacing is, and how the influences vary with prop spacing.
There is a wide range of gaps available, some aircraft, such as the X-57, have a spacing that is very small - the tips are touching.

Image source
So my question is:
What is the aerodynamic influence of prop spacing (to each other and to the engine) in multi-prop engine setups?

@sophit Because the power of electrical motors scales linearly with their mass. For the same power, no matter if you have a single large electrical motor or many small ones, the mass will be similar. For combustion engines it is not the case: one large engine will be much lighter than two small engines with half the power each. Distributed propulsion is not necessarily better, but it has the potential to do things that would not be feasible with a mono-/twin-engine configuration.

@Robe I don't know about the first part of your answer but I guess you're right with the second one: blowing on the wing gives more lift also at low speeds. Thanks for your comment

@sophit Reading these comments made me think about propeller tip speeds and the impact that may have on propulsive efficiency.  Are there advantages to a larger number of smaller propeller disks over a smaller number of larger disks?  To Robe's point, perhaps the weight cost of additional combustion engines justified compromises in propeller disk size, and once that weight cost is removed, smaller is better?

@Robe I suspect that in addition to weight, combustion engines benefit from a smaller count for other reasons as well.  Overhaul costs do not scale linearly with displacement, so 16 tiny engine overhauls are quite likely more expensive than 2 large engine overhauls.  Same would be true of oil changes, spark plugs, magnetos, etc.  There's a lot of combustion engine costs that don't necessarily scale smoothly with engine count.  And I potentially have to route 16 throttle cables, and fuel lines, and primer lines, and and and...

@MaxR: very good points. From the very basic momentum theory we know that the bigger the radius, the more efficient the propeller/rotor is. Many slower rotating propeller can reach the same efficiency but each of them introduces anyway inefficiencies that somehow sum up. On the other hand they are small and light and their rotating speed can be varied in a relative simple way and therefore be used to control the aircraft, see lilium, volocopter and so on. As usual it's a matter of compromises and I personally prefer one very good engine than 64 of them that can get broken.

A major consideration is the single engine performance when one of the engines fail.  If the engines and props are farther away from the aircraft centerline, then the adverse yaw experienced with an engine failure is more extreme.  This would make is much harder to control the aircraft.
In a typical twin-engine propeller driven airplane, when you shut down one engine, the airplane will yaw and roll agresively towards the dead engine.  The pilot has to apply maximum rudder pressure towards the good engine to keep the airplane from completely rolling out of control.  This is one reason the rudder and vertical stabilizer of twins are larger than single engine planes.
Another factor is the structural strength needed to support a heavy piston engine far out from centerline.  So, in general, designers would want the props as close as possible to centerline while taking into account aerodynamic factors such as the prop slipstream and the increased drag caused by the engine nacelles being too close to the fuselage.

Why does drag increase if the engine nacelles are placed close to the fuselage?

I think the question was more about the aerodynamic interference when the rotor are close to each other.


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What is the protruding element on the vertical tail of this Airbus Zero E concept?
The TurboFan concept of Airbus Zero E has a protruding element on the top of its vertical tail, see the image below.

Image source
Zoomed image

Only this variant of the Zero E has it, is it related to the hydrogen turbofan power unit?
It looks like a jettison valve or something like that. I'm wondering because this looks like a drag-sensitive implementation of a release valve.

Perhaps the boil-off valve? But why does it have to protrude out of the vertical stabilizer?

They're definitely vents. Prolly because hydrogen discharge would freeze the tip of the surface?

My guess is that it is the hydrogen emegency venting duct. If the hydrogen starts to boil, it is vented ont the top of the tail because, being lighter than air and being explosive, you want it to stay away from the aircraft. But I don't have any source confirming this.


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Why does elevator still raise and lower the nose when the aircraft is steeply banked?
In a steep turn, the aircraft is basically sideways. However, when executing a steep turn, one still raises and lowers the nose of the aircraft by using the elevator.
Intuitively, it seems to be that in an extreme bank, the elevator and rudder have essentially changed roles, since they have been rotated with respect to their normal orientation. It seems to me that in a steep turn, pulling back on the yoke should tighten the turn. But this seems to not be the case in practice, since getting the nose up in a steep turn still involved just pulling back on the yoke and/or trim.
Why is this the case?

It seems to me like you're switching reference points, saying it does this from the pilot's reference, but looks like that from a ground observer's reference.

@BillDOe Quite possibly my confusion is an issue of reference frames, but I do not see how. Could you elaborate?

Basically Sideways... The FAA considers steep turns as bank angles between 45 and 60 degrees the aircraft is not quite sideways. Both rudder and elevator input is needed to execute properly.

@BillDOe on further reflection, you are correct. I was basing my question on my observation of the attitude indicator - but an upward motion of the nose on the attitude indicator while banked does not mean a purely nose-up response.

@Dave an exaggeration, certainly. The point was that the force exerted by the elevator is not up and down anymore, but has a significant sideways component. But I think I understand better.

As Dave is saying one reason could be that you real bank angle is much lower than perceived, banking 60 degrees feels like really a lot if you are not used to it. Do you know what your real bank angle was when this question arose?

@Martin I was aiming for 45, being new to it I ended up anywhere between 40 and 50 over a few attempts.

At 45 degrees Angle of Bank pulling back will both tighten the turn, and cause the nose to rise equally because you are halfway to 90 degrees.  You are correct though, at much steeper angles of bank the control inputs function the other way:  In a very steep high G turn the angle of bank is used to control altitude, and back pressure is used to control the rate of turn.  You definitely don't want to use rudder in this case though, bad things can happen in an accelerated stall with yaw.

This is what I was looking for. Can you maybe elaborate on what you mean by AOB being used to control altitude, though?

Sure.  Let's say you are striving for a max performance, 4G turn at a target cornering speed, and you are at approximately 75 degrees angle of bank.  If you are holding airspeed and backpressure, (Gs) constant to achieve the best turn performance but notice you are descending, it is because you have over banked a little.  To correct you would lessen your AOB to stop the descent and climb back up.  Conversely if you are climbing in the turn you would overbank a little to descend.

... to take it further, if you held a constant 4Gs and let the airplane roll inverted you'd do a split S.  If you held 4Gs and rolled upright you'd do a loop.  The minor corrections described above just help keep the nose tracking along the horizon for a hard level turn.

First thing to note is the fact that "tightness" (be it radius or rate of turn) of the coordinated, steady, horizontal turn is determined by airspeed and bank angle. Namely radius for banking angle $\alpha$ and airspeed $v$ is
$$
r={v^2\over g \tan\alpha}
$$
and rate of turn (degrees per second)
$$
\hbox{rate}=360° {v\over 2\pi\cdot r} = {360°\over 2\pi}\cdot{g \tan\alpha\over v}
$$
Slowing down takes long, so only straightforward way to increase turn tightness is increasing bank. If you use either elevator or rudder, but try to keep angle the same, you will end with uncoordinated turn at the best case.
Well, uncoordinated turn is not forbidden, so let's analyze this closer.
But first have a look what is needed to feel the turn as tighter. Either you except feeling more G's pushing you into the seat or horizon speeding around before the nose faster. In any case it means that you need more force pointing inside the turn (airspeed unchanged).
Only source of big enough force are wings (any control surfaces are really weak when it comes to generating forces comparable to the airplane weight, the role of control surfaces is slightly turn the airplane in order to make wings to generate lift in right direction). But wing's lift has a direction mostly constant with respect to the airplane and you have limited your maneuvering space by wanting constant bank. Still, increasing lift is your only chance, so let's pull on the yoke.
If we forget about gravity for a moment, pulling on the yoke when 45° banked would result in nice slanted loop (45° inclined w.r.t. horizon and note that angle between wings and horizon would stay 45° all the time). But gravity is here and vertical component of the lift in turn is just compensating it instead of making you fly a loop.
If you pull bit more (say add half of the current lift amount), you will actually ask for flight which is composition of horizontal turn and that slanted loop. Resulting in a "flat" loop inclined 18° above horizon (horizontal component of the force will be 1.5 times original value, new vertical component 0.5 of original horizontal, $\tan^{-1} 1/3 = 18°$).
18° does not seems as a much, but it still means that with 10°/s turn rate the nose will go 3° up in less than second. And this is something, you will notice directly in contrast with less pronounced "more turning".
You can try to compensate with rudder and keep the nose down, but now you are comparing authority of rudder with pitch-wise stability generated by tailplane and wings. Having the same leverage (45° bank) and normal (non-aerobatic) aircraft, it is more or less sure that rudder does not have enough authority and pitch stability wins lifting the nose.
If you would like to continue this "flat loop", you would need to let bank angle to increase slightly. If you compensate instead, you are exchanging even more turning for nose up. And because there is gravity which keeps its orientation in space, your loop won't stay circular anyway. It will end up with some uncoordinated turn soon and if you compensate this, you are back at coordinated turn with original bank angle and therefore the same "tightness".
Your only chance for "tightening" the turn with elevator is having enough rudder authority, so you can hold nose low and fly the skidding turn. But for this you need enough leverage for rudder to overcome pitch-wise stability. Lot of bank in other words. Depending on the particular airplane even 60° does not need to be enough.
Edit: Note that in coordinated turn with constant bank angle the position of nose above or below horizon is not fixed. It will affect power budget of the flight by climbing, sinking and/or changing airspeed, but it has no big and direct effect on turn coordination. So if you struggle to keep you bank angle constant and turn coordinated only thing which can change is just this – angle of nose over horizon. And therefore it is the only change you see when pulling on yoke because you are compensating away the rest of direct effects.

It's because the airplane doesn't actually know it's banked.  To the airplane (and to your body for that matter), gravity is still perpendicular to the wings. It's just that the wings have to lift harder to compensate for having to generate a lateral force to change direction plus still support the airplane's weight in the vertical plane.
And it's elevator that controls how hard the wings have to lift.  If you added several hundred pounds of ballast in straight flight, it's the same thing to the elevator.

Pulling back on the yoke or stick when the aircraft is turning, and therefore banked, will both tighten the turn, and lift the nose relative to the horizon.  The amount of each depends on the bank angle.  In a shallow turn, increasing up elevator will mostly raise the nose and have little effect on turn radius.  In a steep turn, say those over 45 degrees bank, up elevator will act more to tighten the turn, but will still raise the nose, just to a lesser extent.
Only at a bank angle of 90 degrees would the elevator not lift the nose relative to the horizon.  So during steep turns,, the elevator will still control pitch attitude, though to a lesser extent than during shallow turns or level flight.


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What does the flashing green-light signal (return for landing) actually mean to NORDO aircraft?
Studying for the PSTAR I see that a flashing green signal while in the circuit means "return for landing" but this is not actually defined anywhere. What does that mean?
NORDO = "is an aircraft flying without a radio"

What I was taught that this meant was ...
... because they could not communicate with you, but they wanted you to not land on this pass (interval a mess, crossing traffic, winds maybe wrong, a dozen other things that might crop up) it was more or less "the runway will be ready for you if you take another lap in the pattern." This also gives them a bit of time to get all of the other aircraft in the pattern out of the way of the aircraft in distress. (And the steady green ought to confirm that on the next pass, if all things go well).
Tower operators would rather get that aircraft who can't talk to them onto the ground.  What they want to avoid in this case (again, this is what I was taught by a salty old Navy controller when I was in flight school) is that they wanted to avoid the aircraft attempting to land being "spooked" by a red light and thinking they can't land at this field.  And then waving off and flying off in search of another field while still NORDO.
The green light is the 'friendly' light between choices of red and green. Put in the chief's colloquial jargon: "Flashing green lets the pilot know "the runway will be there for you when you come back for your next approach."
The Canadian Aeronautical Information Manual, which has all of the signals to include the two that are relevant to this question, can be found here.
The teaching that I got may no longer be current, but the lights shown to a NORDO airborne aircraft still mean the same thing as they did when I was taught.

Steady Green:  cleared to land
  Flashing Green: return to land (to be followed at the proper time by steady green)

what's the practical difference between flashing green and steady red, then?

You don't know what you'll see on the next pass with a steady red.  I need to dig up our old flying manuals to give you anything more than that.  that may take a day or two.

got it, thanks!

My understanding of light gun signals was taught to me a little differently. For simplicity, I will concentrate on light gun signals for airborne aircraft only.

Alternating Red and Green is exercise extreme caution.
Steady Red is give way to other traffic. Which implies that you should stay in the pattern. But, you are not cleared to land.
Steady Green is a clearance to land.
Flashing Red is a request to leave the airport area. This would imply that you need to remain outside the traffic pattern.
Flashing Green is the signal in question. Based on the definitions of the other signals, this would imply that you should enter or return to the pattern. But, you are still not cleared to land until you get a Steady Green.

So to put it into a nutshell that may be easier to remember:

Green means approval and Red means denial.
Steady means for the runway and Flashing means for the pattern.

"Alternating Red and White" should be "Alternating Red and Green". Also note that this signal is not applicable in Canada (as the question is tagged) but does apply in the USA.

@user2070305 - I thought I had edited this before. Thanks for the catch.

The answers above are pretty much correct, but it is nice to cite sources.
Since the question is relating to studying for a PSTAR and is tagged Canada, first the Canadian sources:
Canada
Aeronautical Information Manual (Canada)
Section 4.2.11 Visual Signals - Aircraft on Ground

4.4.7 Visual Signals - Aircraft in Flight

United States
The Canadian signals are similar, but not exactly the same, as US FAA ATC signals, so adding these for US pilots who find this question
§ 91.125 ATC light signals

In the Airman's Information Manual (US FAA), the signals are described in Section 4-3-13
Differences

Canada doesn't specify the use of alternating red/green signal
USA doesn't specify the use the blinking runway lights nor the firing of the red pyrotechnical light

In the US, to close a runway, a big “X” on the runway is used. In some areas, a large, blinking, lit up “X” is used to mark a temporary closure.

Logically it would seem to mean to fly a circuit around the traffic pattern and watch for a steady green light on the next final approach.

I'm sure some other superior answer will eventually overshadow this one, but it's a start.

@KorvinStarmast -- "I'm sure some other superior answer will eventually overshadow this one"-- as has now happened, nice job on that.


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Can a weather forecast tell me if airplanes will leave long contrails?
Training 7 ball and 5 club juggling requires me to look up into the sky for durations of up to one minute. Frequently, this is very taxing on my eyes and, frustratingly, requires me to wear sunglasses.
Certain atmospheric conditions are much easier on my eyes. In these conditions, the sky is a very deep blue and I can train juggling without sunglasses, even during mid-day. Atmospheric conditions that have good visibility for juggling almost always cause airplanes to leave extremely long contrails.
The Appleman chart shows that contrails form at relatively low temperatures, low relative humidity, and high pressures. Are predictions of these temperatures and pressures available? This information would be helpful in scheduling my juggling training sessions.

Just a note: The linked question features the Appleman chart which has been created for low-bypass-engines. Refer to the new chart by Jeffrey L. Peters for high-bypass-engines. You can find the document here. The chart is on document page 16, respectively PDF page 22.


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Is it possible as a pilot to get the ICAO proficiency language level for a not official language in the ICAO?
I was wondering if it is possible to get the proficiency language level in German for example, as it is not an official language for ICAO. If I would want to apply for a German airline as a pilot, can I get that document, the same as I can get it for Spanish or English?
I am not 100% sure but I think pilots in Germany, Austria and Switzerland can communicate in German so I guess they need those competencies (although when being native you don't have to take the exam)
Maybe it's a dumb question but I haven't found an answer yet.

If you want to fly IFR, or fly to any other country commercially, you need to be proficient in English. VFR in Germany can be conducted in German, but they try to limit it to local flights or emergencies.

As the others pointed out, there's no ICAO language proficiency test for German.
How it works for German at least in Germany is with a special "Radio operator license" (Sprechfunkzeugnis) that's part of (German) pilot licenses. For IFR that's only in English, but for VFR you have the choice of doing it in English, German or both. It's called BZF I (English + German), BZF II (German) or BZF E (English). According to the German Wikipedia, there are similar systems in place in Austria and Switzerland.
Even for VFR in Germany it's totally possible to do it in English, as I believe there are no more German-only speaking airfields. But especially on smaller fields where radio services are provided unpaid by people in their free time, they are a bit happier when using German.
And even on the bigger VFR fields with professional ATC personnel, my experience is that it's sometimes hard for them to remember, which aircraft in the pattern is communicating in English and which in German and then they try to translate position reports they get in one language from one aircraft into traffic advisories for other aircraft in another language. That's why it's sometimes recommended for pilots flying more often VFR in Germany to maybe also do the BZF in German as most VFR traffic is still done in German.

Thank you @Florian:) you solved my doubts. I have looked for BZF and downloaded a book to get a bit into it . I can speak German perfectly but I have no idea about German comms though. Regarding your experience, would you say it is important to get that Sprechfunkzeugnis in order to be able to apply for a German airline?

@Flightfighter No, I wouldn't say so, they should just care about IFR, which is English anyways.

Alright!!! Thank you mate;) safe landings!

@Florian There is no language proficiency test for German, but you can still get LP6 written into your licence/BZF for German if you request it from the license issuing authority. I have BZF I LP5 EN and LP6 DE in my license.

@SentryRaven so then it’s actually possible to get it. But for a no native German speaking person like me, it is necessary to take the BZF exam and then request it from the license issuing authority, in this case the Spanish one, right?

I am not sure you can get DE LP without actually being German. There is no exam and there is no requirements for it. Why the Germans issue it.... no clue really. I had to send a letter with my license and ask to have DE LP6 included cause I am a native speaker. It's purely cosmetic...

@SentryRaven I understand. I thought I could get the DE LP by taking the AZF-prüfung, as I have read in some flight schools websites. Although maybe it's useless for me and having the English one is enough.

English is the only requirement, though in countries where a language is spoken that's also an official language that language may suffice (Spanish in Spain for example).
German may be spoken unofficially in German language countries for non-controlled flight only (though controllers providing flight services to VFR pilots may well communicate in German if called in that language).
As German is not an ICAO language, there is obviously no German ICAO language proficiency test. Maybe EASA's German sister agency has a test of their own, you'd need to contact them or a flight school providing theory training.